Trigonometry Examples

$\frac{x^{2}}{36} + \frac{y^{2}}{20} = 1$

Step 1

Simplify each term in the equation in order to set the right side equal to $1$ . The standard form of an ellipse or hyperbola requires the right side of the equation be $1$ .

$\frac{x^{2}}{36} + \frac{y^{2}}{20} = 1$

Step 2

This is the form of an ellipse. Use this form to determine the values used to find the center along with the major and minor axis of the ellipse.

$\frac{{(x - h)}^{2}}{a^{2}} + \frac{{(y - k)}^{2}}{b^{2}} = 1$

Step 3

Match the values in this ellipse to those of the standard form. The variable $a$ represents the radius of the major axis of the ellipse, $b$ represents the radius of the minor axis of the ellipse, $h$ represents the x-offset from the origin, and $k$ represents the y-offset from the origin.

$a = 6$

$b = 2 \sqrt{5}$

$k = 0$

$h = 0$

Step 4

The center of an ellipse follows the form of $(h, k)$ . Substitute in the values of $h$ and $k$ .

$(0, 0)$

Step 5

Find $c$ , the distance from the center to a focus.

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Step 5.1

Find the distance from the center to a focus of the ellipse by using the following formula.

$\sqrt{a^{2} - b^{2}}$

Step 5.2

Substitute the values of $a$ and $b$ in the formula.

$\sqrt{{(6)}^{2} - {(2 \sqrt{5})}^{2}}$

Step 5.3

Simplify.

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Step 5.3.1

Simplify the expression.

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Step 5.3.1.1

Raise $6$ to the power of $2$ .

$\sqrt{36 - {(2 \sqrt{5})}^{2}}$

Step 5.3.1.2

Apply the product rule to $2 \sqrt{5}$ .

$\sqrt{36 - (2^{2} {\sqrt{5}}^{2})}$

Step 5.3.1.3

Raise $2$ to the power of $2$ .

$\sqrt{36 - (4 {\sqrt{5}}^{2})}$

Step 5.3.2

Rewrite ${\sqrt{5}}^{2}$ as $5$ .

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Step 5.3.2.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{5}$ as $5^{\frac{1}{2}}$ .

$\sqrt{36 - (4 {(5^{\frac{1}{2}})}^{2})}$

Step 5.3.2.2

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$\sqrt{36 - (4 \cdot 5^{\frac{1}{2} \cdot 2})}$

Step 5.3.2.3

Combine $\frac{1}{2}$ and $2$ .

$\sqrt{36 - (4 \cdot 5^{\frac{2}{2}})}$

Step 5.3.2.4

Cancel the common factor of $2$ .

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Step 5.3.2.4.1

Cancel the common factor.

$\sqrt{36 - (4 \cdot 5^{\frac{2}{2}})}$

Step 5.3.2.4.2

Rewrite the expression.

$\sqrt{36 - (4 \cdot 5^{1})}$

Step 5.3.2.5

Evaluate the exponent.

$\sqrt{36 - (4 \cdot 5)}$

Step 5.3.3

Multiply $- (4 \cdot 5)$ .

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Step 5.3.3.1

Multiply $4$ by $5$ .

$\sqrt{36 - 1 \cdot 20}$

Step 5.3.3.2

Multiply $- 1$ by $20$ .

$\sqrt{36 - 20}$

Step 5.3.4

Subtract $20$ from $36$ .

$\sqrt{16}$

Step 5.3.5

Rewrite $16$ as $4^{2}$ .

$\sqrt{4^{2}}$

Step 5.3.6

Pull terms out from under the radical, assuming positive real numbers.

$4$

Step 6

Find the vertices.

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Step 6.1

The first vertex of an ellipse can be found by adding $a$ to $h$ .

$(h + a, k)$

Step 6.2

Substitute the known values of $h$ , $a$ , and $k$ into the formula.

$(0 + 6, 0)$

Step 6.3

Simplify.

$(6, 0)$

Step 6.4

The second vertex of an ellipse can be found by subtracting $a$ from $h$ .

$(h - a, k)$

Step 6.5

Substitute the known values of $h$ , $a$ , and $k$ into the formula.

$(0 - (6), 0)$

Step 6.6

Simplify.

$(- 6, 0)$

Step 6.7

Ellipses have two vertices.

${Vertex}_{1}$ : $(6, 0)$

${Vertex}_{2}$ : $(- 6, 0)$

${Vertex}_{1}$ : $(6, 0)$

${Vertex}_{2}$ : $(- 6, 0)$

Step 7

Find the foci.

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Step 7.1

The first focus of an ellipse can be found by adding $c$ to $h$ .

$(h + c, k)$

Step 7.2

Substitute the known values of $h$ , $c$ , and $k$ into the formula.

$(0 + 4, 0)$

Step 7.3

Simplify.

$(4, 0)$

Step 7.4

The second focus of an ellipse can be found by subtracting $c$ from $h$ .

$(h - c, k)$

Step 7.5

Substitute the known values of $h$ , $c$ , and $k$ into the formula.

$(0 - (4), 0)$

Step 7.6

Simplify.

$(- 4, 0)$

Step 7.7

Ellipses have two foci.

${Focus}_{1}$ : $(4, 0)$

${Focus}_{2}$ : $(- 4, 0)$

${Focus}_{1}$ : $(4, 0)$

${Focus}_{2}$ : $(- 4, 0)$

Step 8

Find the eccentricity.

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Step 8.1

Find the eccentricity by using the following formula.

$\frac{\sqrt{a^{2} - b^{2}}}{a}$

Step 8.2

Substitute the values of $a$ and $b$ into the formula.

$\frac{\sqrt{{(6)}^{2} - {(2 \sqrt{5})}^{2}}}{6}$

Step 8.3

Simplify.

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Step 8.3.1

Simplify the numerator.

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Step 8.3.1.1

Raise $6$ to the power of $2$ .

$\frac{\sqrt{36 - {(2 \sqrt{5})}^{2}}}{6}$

Step 8.3.1.2

Apply the product rule to $2 \sqrt{5}$ .

$\frac{\sqrt{36 - (2^{2} {\sqrt{5}}^{2})}}{6}$

Step 8.3.1.3

Raise $2$ to the power of $2$ .

$\frac{\sqrt{36 - (4 {\sqrt{5}}^{2})}}{6}$

Step 8.3.1.4

Rewrite ${\sqrt{5}}^{2}$ as $5$ .

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Step 8.3.1.4.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{5}$ as $5^{\frac{1}{2}}$ .

$\frac{\sqrt{36 - (4 {(5^{\frac{1}{2}})}^{2})}}{6}$

Step 8.3.1.4.2

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$\frac{\sqrt{36 - (4 \cdot 5^{\frac{1}{2} \cdot 2})}}{6}$

Step 8.3.1.4.3

Combine $\frac{1}{2}$ and $2$ .

$\frac{\sqrt{36 - (4 \cdot 5^{\frac{2}{2}})}}{6}$

Step 8.3.1.4.4

Cancel the common factor of $2$ .

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Step 8.3.1.4.4.1

Cancel the common factor.

$\frac{\sqrt{36 - (4 \cdot 5^{\frac{2}{2}})}}{6}$

Step 8.3.1.4.4.2

Rewrite the expression.

$\frac{\sqrt{36 - (4 \cdot 5^{1})}}{6}$

Step 8.3.1.4.5

Evaluate the exponent.

$\frac{\sqrt{36 - (4 \cdot 5)}}{6}$

Step 8.3.1.5

Multiply $- (4 \cdot 5)$ .

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Step 8.3.1.5.1

Multiply $4$ by $5$ .

$\frac{\sqrt{36 - 1 \cdot 20}}{6}$

Step 8.3.1.5.2

Multiply $- 1$ by $20$ .

$\frac{\sqrt{36 - 20}}{6}$

Step 8.3.1.6

Subtract $20$ from $36$ .

$\frac{\sqrt{16}}{6}$

Step 8.3.1.7

Rewrite $16$ as $4^{2}$ .

$\frac{\sqrt{4^{2}}}{6}$

Step 8.3.1.8

Pull terms out from under the radical, assuming positive real numbers.

$\frac{4}{6}$

Step 8.3.2

Cancel the common factor of $4$ and $6$ .

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Step 8.3.2.1

Factor $2$ out of $4$ .

$\frac{2 (2)}{6}$

Step 8.3.2.2

Cancel the common factors.

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Step 8.3.2.2.1

Factor $2$ out of $6$ .

$\frac{2 \cdot 2}{2 \cdot 3}$

Step 8.3.2.2.2

Cancel the common factor.

$\frac{2 \cdot 2}{2 \cdot 3}$

Step 8.3.2.2.3

Rewrite the expression.

$\frac{2}{3}$

Step 9

These values represent the important values for graphing and analyzing an ellipse.

Center: $(0, 0)$

${Vertex}_{1}$ : $(6, 0)$

${Vertex}_{2}$ : $(- 6, 0)$

${Focus}_{1}$ : $(4, 0)$

${Focus}_{2}$ : $(- 4, 0)$

Eccentricity: $\frac{2}{3}$

Step 10